Question

A thin converging lens of focal length $ f = 25\, cm $ forms the image of an object on a screen placed at a distance of $ 75\, cm $ from the lens. The screen is moved closer to the lens by a distance of $ 25\, cm $ . The distance through which the object has to be shifted, so that its image on the screen in sharp again is

• A. 37.5 cm
• B. 16.25 cm
• C. 12.5 cm
• D. 13.5 cm

Answer 1

Answer: (C)

According to the first condition

$ f =25 \,cm , v=75 \,cm $
$u =? $
$ \frac{1}{f} =\frac{1}{v}-\frac{1}{u} $
$ \frac{1}{25} =\frac{1}{75}-\frac{1}{u}$
$\frac{1}{u} =\frac{1}{75}-\frac{1}{25} $
$ \frac{1}{u} =\frac{1-3}{75} $
$ u =-\frac{75}{2}=-37.5 \,cm$
According to the second condition

$v_{1}=50 \, cm , f =25 \, cm , u_{1}=? $
$\frac{1}{f} =\frac{1}{v_{1}}-\frac{1}{u_{1}} $
$\frac{1}{25} =\frac{1}{50}-\frac{1}{u_{1}} $
$\Rightarrow \frac{1}{u_{1}}=\frac{1}{50}-\frac{1}{25}$
$\Rightarrow \frac{1}{u_{1}}=\frac{1-2}{50}$
$u_{1}=-50 \,cm$
So, the screen is sharp again is
$\Delta u =u_{1}-u $
$=50-37.5$
$=12.5 \, cm $